Question

hands-on labs periodic table of elements h pb h₂so₄ 74 g/mol ca(oh)₂ 98 g/mol cuso₄·5h₂o 212 g/mol k₃po₄

Explanation:

Step1: Calculate molar mass of \( H_2SO_4 \)

Atomic masses: \( H = 1 \), \( S = 32 \), \( O = 16 \).
Molar mass \( = 2\times1 + 32 + 4\times16 = 2 + 32 + 64 = 98 \, \text{g/mol} \).

Step2: Calculate molar mass of \( Ca(OH)_2 \)

Atomic masses: \( Ca = 40 \), \( O = 16 \), \( H = 1 \).
Molar mass \( = 40 + 2\times(16 + 1) = 40 + 34 = 74 \, \text{g/mol} \).

Step3: Calculate molar mass of \( CuSO_4 \cdot 5H_2O \)

Atomic masses: \( Cu = 64 \), \( S = 32 \), \( O = 16 \), \( H = 1 \).
Molar mass \( = 64 + 32 + 4\times16 + 5\times(2\times1 + 16) = 64 + 32 + 64 + 90 = 250 \, \text{g/mol} \) (note: 250, likely a typo in 255? But following calculation: 64+32=96; 416=64; 96+64=160; 5(2+16)=90; 160+90=250).

Step4: Calculate molar mass of \( K_3PO_4 \)

Atomic masses: \( K = 39 \), \( P = 31 \), \( O = 16 \).
Molar mass \( = 3\times39 + 31 + 4\times16 = 117 + 31 + 64 = 212 \, \text{g/mol} \).

Answer:

• \( H_2SO_4 \) → 98 g/mol
• \( Ca(OH)_2 \) → 74 g/mol
• \( CuSO_4 \cdot 5H_2O \) → 250 g/mol (or 255? If Cu=63.5, S=32, O=16, H=1: 63.5+32+64 + 5*(18)=159.5+90=249.5≈250 or 255? Maybe Cu=64, S=32, O=16, H=1: 64+32+64 + 90=250. If 255, maybe Cu=65? Anyway, per standard: \( CuSO_4 \cdot 5H_2O \) is ~250 g/mol (exact 249.685).
• \( K_3PO_4 \) → 212 g/mol

So matches:
\( H_2SO_4 \) - 98 g/mol
\( Ca(OH)_2 \) - 74 g/mol
\( K_3PO_4 \) - 212 g/mol
\( CuSO_4 \cdot 5H_2O \) - 250 (or 255) g/mol (assuming 250 as per calculation).

(Note: The left boxes: 74→\( Ca(OH)_2 \), 98→\( H_2SO_4 \), 212→\( K_3PO_4 \), 250→\( CuSO_4 \cdot 5H_2O \))